Solution Manual For Electrical Engineering 7th Edition by Hambley Chapter 17 Not Included PDF
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CHAPTER 1 Exercises E1.1
Charge = Current Ti Time me = (2 A) (10 s) = 20 C
E1.2
i (t )
E1.3
Because i 2 has a positive value, positive charge moves in the same direction as the reference. Thus, positive charge moves downward in element C .
dq (t ) d (0.01sin(20 0t) 0.01 200cos(200t ) 2cos(200t ) A dt dt
Because i 3 has a negative value, positive charge moves in the opposite direction to the the reference. Thus positive charge mo moves ves upward in element E . E1.4
Energy = Charge Voltage = (2 C) (20 V) = 40 J is positive, the positive terminal is a and and the negative Because v aab b is terminal is b . Thus the charge moves ffrom rom the negative terminal to tthe he positive terminal, and energy is removed from the circuit element.
E1.5
i aab enters terminal a . Furthermore, v aab is positive at terminal a . Thus b enters b is
the current enters the positive reference, and we have the passive reference configuration. E1.6
(a) p a (t ) v a (t ) i a (t ) 20t 2 10
10
10
20 20t 3 t 3 6667 J w a p a (t )dt 20t dt 3 3 0 0 0 (b) Notice that the rreferences eferences are op opposite posite to the passive sign convention. Thus we have: 2
200 0 p b (t ) v b (t )i b (t ) 20t 20 10
10
0
0
10
200 0)dt 200 0t 0 1000 J w b p b (t )dt (20t 20 10t 2 20
1
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E1.7
(a) Sum of curren currents ts leaving = Sum of currents en entering tering i a = 1 + 3 = 4 A a = (b) 2 = 1 + 3 + i bb
(c) 0 = 1 + i c + 4 + 3 c +
i bb = = -2 A
i c = -8 A c =
E1.8
Elements A and and B are are in serie series. s. Also, elem elements ents E , F , and G are in series.
E1.9
Go clockwise around the loop consisting of elements A , B , and C : = 8 V = 0 v c -3 - 5 +v cc = c = Then go clockwise around the loop composed of elements C , D and and E : - v c - (-10) + v ee = 0 v e c e = -2 V
E1.10
Elements E and and F are are in parallel; elements A and and B are are in series.
E1.11
The resistance of a wire is given by R
ρL
A
. Using A d 2 / 4 and
substituting values, we have:
1.12 106 L 9.6 3 2 (1.6 10 ) / 4
L = = 17.2 m
I V / R 12 120 0 / 14 144 4 0.83 833 3 A
144 4 R V 2 / P 14
E1.12
P V 2 R
E1.13
P V R V PR 0.25 1000 15.8 V / 1000 15.8 mA I V / R 15.8
E1.14
Using KCL at the top node of the circuit, we have i 1 = i 2. Then, using KVL going clockwise, we have -v 1 - v 2 = 0; but v 1 = 25 V, so we have v 2 = -25 V. Next we have i 1 = i 2 = v 2/R = = -1 A. Finally, we have 1i 1 (25) (1) 25 W. P R v 2i 2 (25 ) (1) 25 W and P s v
E1.15
At the top node we have i RR = = i s = 2A. By Ohm’s law we have v RR = = Ri RR = = 80 V. By KVL we have v ss = = v RR = = 80 V V.. Then p s s = = -v s i s = -160 W (the minus sign si is due to the fact that the references for v s s and and i s s are are opposite to the 160 W. passive sign configuration). Also we have P R v R i R 160
2
2
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Problems P1.1
Broadly, the two objectives of electrical systems are: 1. To gather, store, process, transport and present information. 2. To distribute, store, and convert energy between various forms.
P1.2
Four reasons that non-electrical engineering majors need to learn the fundamentals of EE are: 1. To pass the Fundamentals ooff Engin Engineering eering Exam. 2. To be able to lead in the design of systems that contain electrical/electronic electrical/electron ic elements. 3. To be able to operate operate and maintain systems that contain electrical/electronic electrical/electron ic functional blocks. 4. To be able to communicate communicate effec effectively tively with electrical electrical engineers.
P1.3
Eight subdivisions of EE are: 1. 2. 3. 4. 5. 6. 7. 8.
Communication systems. Computer systems. Control systems. Electromagnetics. Electronics. Photonics. Power systems. Signal Processing.
P1.4
Responses to this question are varied.
P1.5
(a) Electrical current is the time rate of flow of net charge through a conductor or circuit circuit elemen element. t. Its units are ampe amperes, res, which are eequivalent quivalent to coulombs per second. (b) The voltage between two points in a circuit is the amoun amountt of energy transferred per unit unit of charge movin movingg between the points. Voltage has units of volts, which are equivalent to joules per coulomb.
3
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(c) The current current through an open switch is zero. The voltage across the switch can be any value depending on the circuit. (d) The voltage across across a closed switch is zzero. ero. The current th through rough the switch can be any value depending of the circuit. (e) Direct current is constant in magnitude and direction with respect to time. (f) Alternating current current varies eithe eitherr in magnitude or direction with with time.
P1.6
(a) A conductor is anagolous to a frictionless pipe. (b) An open switch is anagolous to a closed valve. (c) A resistance is anagolous to a constriction in a pipe or to a pipe with friction. (d) A battery is analogous to a pump.
1 coulomb/s 6.25 1018 19 1.60 10 coulomb/el ectron
P1.7
Electrons pe perr second
P1.8*
The reference direction for i ab points from a to to b. Because i ab has a negative value, the current is equivalent to positive charge moving opposite to the reference reference direction. direction. Finally since elec electrons trons have negative charge, they are moving in the reference direction (i.e., from a to to b ). ). For a constant (dc) current, charge equals current times the time interval. Thus, Q (5 A) (3 s) 15 C.
P1.9
The positive reference for v is at the head of the arrow, which is terminal a . The positive reference for v ba ba is terminal b . Thus, we have v ba v 12 V. Also, i is the current entering terminal a , and i bba is the a is current leaving terminal a . Thus, we have i i ba 2 A. Thus, current enters the positive reference and energy is being delivered to the device.
P1.10
To stop current flow, we break contact between the conducting parts of the switch, and we say that the switch is open. The corresponding fluid
4
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analogy is a valve that does no nott allow fluid to pass through. This corresponds to a closed closed valve. Thus, a closed valve is analo analogous gous to an open switch. P1.11* i t
P1.12
dq t d 2 3 t 3 A dt dt
(a) The sine function completes one cycle for each 2 radian increase in the angle. angle. Because the angle is 20 200 0 t , one cycle is completed for each time interval interval of 0.01 s. The sketch is:
(b) (b) Q
0.00 005 5
0.00 005 5
0
0
i (t )dt
0.0318 C
0.01
(c) (c) Q
0.01
0.01
200 0 t )d t (10 / 20 200 0 ) cos(20 200 0 t ) 0 0 i (t )dt 0 10 sin(20
0 C
P1.13*
0.005 005
200 0 t ) dt (10 / 20 200 0 ) cos(20 200 0 t ) 0 10 sin(20
t Q i (t )dt 2e dt 2e t | 0 2 coulombs
0
0
dq (t ) d 3 3e 2t 6e 2t A dt dt
P1.14
i (t )
P1.15
The number of electrons passing through a cross section of the wire per second is 15 N 9. 37 375 5 1019 electrons/ second 19 1.6 10
5
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The volume of copper containing this number of electrons is
9.37 375 5 1019 volume 9.37 375 5 1010 m3 29 10 The cross sectional area of the wire is 2 d
A
4
3.30 3011 10 6 m2
Finally, the average velocity of the electrons is volume u 0.2840 mm/s A P1.16*
The charge flowing through the battery is 432 2 103 coulombs Q (5 amperes ) (24 3600 seconds ) 43 and the stored energy is Energy QV (43 432 2 103 ) (12) 5.184 184 106 jou joules les (a) Equating gravitational potential energy, which is mass times height times the acceleration due to gravity, to the energy stored in the battery and solving for the height, we have Energy 5.184 184 106 h 17.6 km mg 30 9.8 (b) Equating kinetic energy to stored energy and solving for velocity, we have 2 Energy 587 7.9 m/s v 58 m (c) The energy density of the battery is 5.18 184 4 106 172 172 .8 103 J/kg 30 which is about 0.384% of the energy density of gasoline.
P1.17
) (20 seconds ) 40 coulombs Q current time (2 amperes Energy QV (20 20) ) (10) 20 200 0 jou joules les Because i bba is positive, if the current were carried by positive charge it a is would be entering terminal b . Electrons enter terminal a . The energy is
taken from the element.
6
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P1.18
Thee electron gains 1.6 1019 9 14 .4 1019 jou Th joules les
P1.19*
000 0 seconds ) 1.8 105 coulombs Q current time (5 amperes ) (36,00 Energy QV (1.8 105 ) (12) 2.16 106 jou joules les
P1.20
If the current is referenced to flow into the positive reference for the voltage, we say that we have the passive re reference ference con configuration. figuration. Using double subscript notation, if the order of the subscripts are the same for the current and voltage, we have a passive refe reference rence configuration. configuration.
P1.21*
i aa = 30 W (a) P -v aa i
Energy is being absorbed by the element.
(b) P v bb i i bb = 30 W
Energy is being absorbed by the element.
(c) P -v DE iED D = -60 W DE i E
Energy is being supplied by the element.
P1.22
The amount of energy is W QV (3 C) (1 (10 0 V) 30 J. Because the reference polarity is positive at terminal a and and the voltage value is negative, terminal b is is actually the positive te terminal. rminal. Because the ccharge harge moves from the negative terminal to the positive terminal, energy is removed from the device.
P1.23*
Q w V (600 J) (1 (12 2 V) 50 C .
To increase the chemical energy stored in the battery, positive charge should move from the positive terminal to the negative terminal, in other words from a to to b . Electrons move from b to a . P1.24
p t v t i t 20e t W
Energy p (t )dt joules les 20e t | 0 20 jou
0
The element absorbs the energy.
7
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P1.25
(a) p (t ) v ab i ab 50 sin(20 200 0 t ) W
(b) (b) w
0.005
0.005
0
0
p (t )dt
P1.26*
0.01
0.01
0
0
0.01
200 0 t ) dt (50 / 20 200 0 ) cos(20 200 0 t ) 0 50 sin(20
0 J
60 500 0 kW kWh h Energy Cost 50 Rate 0.12 $/kWh P
Energy 50 500 0 kW kWh h 694.4 W Time 30 24 h
Reduction P1.27
0.1592 J
(c) (c) w p (t )dt
0.005
200 0 π t ) dt (50 / 20 200 0 π ) cos(20 200 0 π t ) 0 50 sin(20
I
694 4 .4 P 69 787 7 A 5.78 120 120 V
60 100 % 8.64% 100 694 69 4.4
(a) P 60 W delivered to element A. P
(b) 60W Wdelivered taken from . . (c) P 60 to element elementAA P1.28*
element A . (a) P 60 W taken from (b) P 60 W delivered to element A . (c) P 60 W taken from element A .
P1.29
The power that can be delivered by the cell is p vi 0.12 W. In 75 hours, the energy delivered is W p pT T 9 W hr 0.009 kWhr. Thus the unit cost of the energy is Cost (0.50) /( 0.00 009 9) 55.56 /kWhr which is 463 times the typical cost of energy from electric utilities.
8
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P1.30
The current supplied to the electronics is i p /v 50 / 12.6 3.96 968 8 A. The ampere-hour rating of the battery is the operating time to discharge the battery battery multiplied by the current. Thus, the operating ttime ime is T 10 100 0 / i 25.2 hours. The energy delivered by the battery is wh 1.26 kWh. Neglecting the cost of W p pT T 50(25.2) 1260 recharging, the cost of energy for 300 discharge cycles is 300 0 1.26) 0.1984 /kWh. Cost 75 /(30
P1.31
A node is a point that joins two oorr more circuit circuit elements. All points joined by ideal conductors conductors are electrically electrically equivalent. Thus, there are four nodes in the circuit at hand:
P1.32
The sum of the currents entering a node equals the sum of the currents leaving.
P1.33
The currents in series-connected elements are equal.
P1.34
For a proper fluid analogy to electric circuits, the fluid must be incompressible. Otherwise the fluid flow rate out ooff an element could be more or less than the inward flow. Similarly the pipes must be inelastic so the flow rate is the same at a t all points along each pipe.
P1.35*
Elements A and B are are in serie series. s. Also, elem elements ents E and and F are in series.
9
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P1.36
(a) Elements C and and D are in series. (b) Because elements C and and D are are in series, the currents are equal in magnitude. However, be because cause the referenc referencee directions ar aree opposite, the algebraic signs of the current values are opposite. Thus, we have i c i d . (c) At the node join joining ing elemen elements ts A , B , and C , we can write the KCL
i b i a i c 3 1 4 A . Also we found earlier that equation i d i c 1 A. P1.37*
At the node joining elements A and B, we have i a i b 0. Thus, i a 2 A. For the node at the top end of element C, we have ha ve i b i c 3 . Thus, i c 1 A . Finally, at the top right-hand corner node, we have and B are in series. 3 i e i d . Thus, i d 4 A . Elements A and
P1.38*
We are given i a 2 A, i b 3 A, i d 5 A, an and d i h 4 A. Applying KCL, we fin find d i e i c i h 5 A i g i f i h 7 A
i c i b i a 1 A i f i a i d 3 A P1.39
We are given i a 1 A, i c 3 A , i g 5 A, and and i h 1 A. Applying KCL, we find i e i c i h 4 A i f i g i h 6 A
i b i c i a 2 A i d i f i a 7 A P1.40
If one travels around a closed path adding the voltages v oltages for which one enters the positive reference and subtracting the voltages for which one enters the negative reference, the total is zero.
P1.41
(a) Elements A and and B are are in parallel. (b) Because elements A and and B are are in parallel, the voltages are equal in magnitude. However because the reference polarities are opposite, opposite, the algebraic signs of the voltage values are opposit opposite. e. Thus, we have v a v b . (c) Writing a KVL equation while going clockwise around the loop composed of elements A, C and and D , we obtain v a v d v c 0. Solving for v c and substituting values, we find v c 7 V. Also we have v b v a 2 V.
10
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P1.42*
Summing voltages for the lower left-hand loop, we have 5 v a 10 0, which yields v a 5 V. Then for the top-most loop, we have a 0, which yields v c 10 V. Finally, writing KCL around the v c 15 v outside loop, we have 5 v c v b 0, which yields v b 5 V.
P1.43
We are given v a 5 V, v b 7 V, v f 10 V, and and v h 6 V. Applying KVL, we find v c v a v f v h 1 V v d v a v b 12 V v g v e v h 2 V v e v a v c v d 8 V v b v c v e 7 V
P1.44*
Applying KCL and KVL, we have i c i a i d 1 A v b v d v a 6 V The power for each element is a i a 20 W P A v P C v c i c 4 W Thus, P A P B P C P D 0
i b i a 2 A v c v d 4 V P B v b i b 12 W P D v d i d 4 W
P1.45
(a) In Figure P1.28, elements C , D , and E are are in parallel. (b) In Figure P1.33, no element is in parallel with another element. (c) In Figure P1.34, elements C and and D are are in parallel.
P1.46
The points and the voltages specified in the problem statement are:
Applying KVL to the loop abca , substituting values and solving, we obtain: v ab v cb v ac 0 v ac 10 V 5 15 v ac 0
11
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Similiarly, applying KVL to the loop abcda , substituting values and solving, we obtain: v ab v cb v cd v da 0 5 15 v cd 10 0 v 20 V cd
P1.47
(a) The voltage between any two points of an ideal conductor is zero regardless of the current flowing. (b) An ideal voltage source maintains a specified voltage across its terminals. (c) An ideal current source maintains a specified current through itself.
P1.48
Four types of controlled sources and the units for their ga gain in constants are: 1. 2. 3. 4.
P1.49
Voltage-controlled voltage sources. V/V or unitless. Voltage-controlled current sou sources. rces. A/V or siemens. Current-controlled Current-controlled voltage sources. V/A or ohms. Current-controlled Current-controlled current sources. A/A or unitless.
Provided that the current reference points into the positive voltage reference, the voltage across a resistance equals the current through the resistance times the resistance. On the other hand, if the current reference points into the negative the voltage equals the negative of the product of the voltage currentreference, and the resistance.
P1.50*
12
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P1.51
P1.52
The resistance of the copper wire is given by R Cu Cu L A , and the resistance of the tungsten wire is R W W L A . Taking the ratios of the respective sides of these equations yields R W R Cu W Cu . Solving for R W and substituting values, we have R W R Cu ρW
ρCu
(0.5) (5.44 10-8 ) (1.72 10 8 ) 1.58 Ω P1.53
P1.54
P1.55*
R
V 1 2 P 1
V 2
100 100 2 100 100 100 100
2
902 P 2 R 100 100 81 W fo forr a 19% 19% reduction in power
13
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P1.56
The power delivered to the resistor is p (t ) v 2 (t ) / R 2.5 exp( 4t ) and the energy delivered is
2.5 2.5 0.62 exp( 4t ) 625 5 J w p (t )dt 2.5 exp( 4t )dt 0 0 4 0 4 P1.57
The power delivered to the resistor is p (t ) v 2 (t ) / R 2.5 sin2 ( 2 t ) 1.25 1.25 cos( 4 t ) and the energy delivered is 10
10
10
1.25 sin(4 t ) 12.5 J w p (t )dt 1.25 1.25 cos( 4 t )dt 1.25t 4 0 0 0 P1.58
Equation 1.10 gives the resistance as R ρL A (a) Thus, if the length of the wire is doubled, the resistance doubles to 1 . (b) If the diameter of the wire is doubled, the cross sectional area A is is increased by a factor factor of four. Thus, the resistance is dec decreased reased by a factor of four to 0.125 .
P1.59
(a) The voltage across the voltage source is 10 V independent of the current. Thus, we have v 10 which plots as a vertical line in the v i plane. (b) The current source has i 2 independent of v , which plots as a horizontal line in the v i plane. plane. (c) Ohm's law gives i v /5. /5. (d) Applying Ohm's law and KVL, we obtain 2. to i 0 .2v
v
5 i 10 which is equivalent
(e) Applying KCL and Ohm's law, we obtain obtain v 5 i 10 which is equivalent to
i 0 .2v 2.
14
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The plots for all five parts are shown. (Parts d and e have the ssame ame plot.)
P1.60*
(a) Not contradicto contradictory. ry. (b) A 2-A current source in series with a 3-A current source is contradictory because the currents in series elements must be equal. (c) Not contradicto contradictory. ry. (d) A 2-A current source in series with an open circuit is contradictory because the current through a short s hort circuit is zero by definition and currents in series elements must be equal. (e) A 5-V voltage source in parallel with a short circuit is contradictory because the voltages across parallel elements must be equal and the voltage across a short circuit is zero by definiton.
P1.61
The power for each element element is 20 W. The current sou source rce delivers power power and the voltage source absorbs it.
15
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P1.62*
As shown above, the 2 A current circulates clockwise through all three elements in the circuit circuit.. Applying KVL, we have v c v R 10 5i R 10 20 V P current source v c i R 40 W. Thus, the current source delivers power.
resistor sistor absorbs power. P R (i R )2 R 22 5 20 W. The re bsorbs power. power. P voltage source 10 i R 20 W. The voltage source aabsorbs P1.63
This is a parallel par allel circuit and the voltage across each element is 10 V positive at the top end. end. Thus, the curren currentt through the resisto resistorr is 10 V 2A i R 5 Applying KCL, we find that the current through the voltage source is zero. Computing powe powerr for each element, w wee find P current source 20 W Thus, the current source delivers power. P R (i R )2 R 20 W P voltage source 0
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Full file at https://testbanku. https://testbanku.eu/Solution-Manualeu/Solution-Manual-for-Electrical-Engi for-Electrical-Engineering-7th-Edition neering-7th-Edition-by-Hambley-Chap -by-Hambley-Chapter-17-not-include ter-17-not-include
P1.64*
Applying Ohm's law, we have v 2 5 1 A 5 V . However,v 2 is th thee voltage across all three resistors that are in parallel. Thus, v v i 3 2 1 A , and i 2 2 0.5 A . Applying KCL, we have 5 10 i 1 i 2 i 3 1 2.5 A . By Ohm's law: v 1 5i 1 12.5 V . Finally using KVL, 2 17.5 V . we havev x v 1 v
P1.65
Ohm’s law for the 5- resistor yields: i 1 15 / 5 3 A. Then for the 10- v i v v 15 45 V. 1 1 2 1 resistor, we have we4.have 10 we 30 V. Using / 10 Then applying Ohms law, obtain 5 A. Finally applying i 2 v 2KVL, KCL, we have I x i 1 i 2 7.5 A. P1.66
(a) The 3- resistance, the 2- resistance, and the voltage source V xx are are in series. (b) The 6- resistance and the 12- resistance are in parallel. (c) Refer to the sketch of the circuit. Applying Ohm's law to the 1212 - resistance, we determine that v 6 V. Then, applying Ohm's law to the 1 6- resistance, we have i 1 1 A. Next, KVL yields i 2 1.5 A. Continuing,
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Full file at https://testbanku. https://testbanku.eu/Solution-Manualeu/Solution-Manual-for-Electrical-Engi for-Electrical-Engineering-7th-Edition neering-7th-Edition-by-Hambley-Chap -by-Hambley-Chapter-17-not-include ter-17-not-include
we use Ohm's law to find that v 2 3 V and v 3 4.5 V. Finally, applying KVL, we have V x v 3 v 1 v 2 13.5 V.
P1.67
First, we have i o P o / 8 1 A.
Applying Ohm's law and KVL to the right-hand loop we have 1000v in 2i o 8i o from which we determine that v in 10 mV. 4 Then, i in v in / 10 1 A , and finally we have V x 5000i in 10000 i in 15 mV. P1.68
(a) No elements are in series.
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Full file at https://testbanku. https://testbanku.eu/Solution-Manualeu/Solution-Manual-for-Electrical-Engi for-Electrical-Engineering-7th-Edition neering-7th-Edition-by-Hambley-Chap -by-Hambley-Chapter-17-not-include ter-17-not-include
and the 2- resistor are in parallel. Also, the 6- resistor and the (b) R x x and the 4- resistor are are in paralle parallel.l. Thus, the voltages across the parallel parallel elements are the same as labeled in the figure. (c)
v 1 6 V i 4 v 1 / 4 1.5 A v x 10 v 1 4 V /2 2A i 2 v x i s i 2 1 1 A i x i 4 i s 0.5 A R x v x / i x 8
P1.69
i 12 v / 12 i 6 v / 6 v 12 V i 12 1 A
P1.70*
(a) Applying KVL, we have 10 v 667 7 V x 5v x , which yields v x 10 / 6 1.66
i 12 i 6 v / 12 v / 6 3 i 6 2 A
(b) i x v 0.5556 A x / 3
556 6 W. (This represen (c) P voltage source 10i x 5.55 represents ts powe powerr delivered by the voltage source.) 926 6 W (absorbed) P R 3 (i x )2 0.92 P controlled source 5v x i x 4.63 W (absorbed) P1.71
Applying KVL around the periphery of the circuit, we have
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Full file at https://testbanku. https://testbanku.eu/Solution-Manualeu/Solution-Manual-for-Electrical-Engi for-Electrical-Engineering-7th-Edition neering-7th-Edition-by-Hambley-Chap -by-Hambley-Chapter-17-not-include ter-17-not-include
18 v x 2v x 0, which yields v x 6 V. V. Then we have v 12 2 v x 12 V. 3 A. Then Using Ohm’s law we obtain i 12 v 12 / 12 1 A and i x v x / 2 KCL applied to the node at the top of the 12- resistor gives i x i 12 i y
which yields i y 2 A. P1.72
Consider the series series combination shown below oonn the left. Because the current for series elements must be the same and the current for the current source is 2 A by definition, the current flowing from a to to b is is 2 A. Notice that the current is not affected by the 10-V source in series. Thus, the series combination is equivalent to a simple current source as far as anything connected to terminals a and b is is concerned.
P1.73
Consider the parallel combination combination shown b below. elow. Because the voltage for parallel elements must be the same, the voltage v aab must be 10 V. Notice b must is not affected affected by the current source. Thus, the parallel that v aab b is combination is equivalent to a simple voltage source as far as anything connected to terminals a and b is is concerned.
P1.74
1 v 2 (a) 10 v (b) v 1 15i v 2 5i (c) 10 15 i 5i i 0.5 A (d) P voltage source 10i 5 W. (Power delivered by th thee source.) P 15 15i 2 3.75 W (absorbed) P 5i 2 1.25 W (absorbed) 5
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Full file at https://testbanku. https://testbanku.eu/Solution-Manualeu/Solution-Manual-for-Electrical-Engi for-Electrical-Engineering-7th-Edition neering-7th-Edition-by-Hambley-Chap -by-Hambley-Chapter-17-not-include ter-17-not-include
P1.75*
1 3 A i s v x / 2
v x (4 ) (1 A) 4 V
Applying KVL around the outside of the circuit, we have: v s 3 i s 4 2 15 V
P1.76
x
i 30 V/15 2A Applying KCL for the node at the top end of the controlled current source: i s i x / 2 i x i x / 2 1 A The source labled i ss is is an independent independent current source. The source labe labeled led i x / 2 is a current-cont current-controlled rolled current source. x/2 P1.77
Applying Ohm's law and KVL, we have 20 10i x 5i x . Solving, we obtain i x 4 A. The source labeled 20 20 V is an independent independent voltage source source.. The source is a current-con current-controlled trolled voltage source. labeled 5i xx is
Practice Test T1.1
(a) 4; (b) 7; (c) 16; (d) 18; (e) 1; (f) 2; (g) 8; (h) 3; (i) 5; (j) 15; (k) 6; (l) 11; (m) 13; (n) 9; (o) 14.
T1.2
(a) The current I ss 3 A circulates clockwise through the elements entering the resistance at the negative reference for v RR . Thus, we have 6 V. R I ss R R v R
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Full file at https://testbanku. https://testbanku.eu/Solution-Manualeu/Solution-Manual-for-Electrical-Engi for-Electrical-Engineering-7th-Edition neering-7th-Edition-by-Hambley-Chap -by-Hambley-Chapter-17-not-include ter-17-not-include
enters the negative reference for V s s, we have P VV V s s II s s (b) Because I ss enters 30 W. Because the result is negative, the voltage source is delivering energy. (c) The circuit has three nodes, one on each of the top corners and one along the bottom of the circuit. (d) First, must findshown: the voltage v I across the current source. We choose thewereference
Then, going around the circuit counterclockwise, we have v I V s v R 10 6 4 V. Next, the power s v R 0 , which yields v I V for the current source is P I I s v I 12 W. Because the result is positive, the current source is absorbing energy. Alternatively, we could compute the power delivered to the resistor as p ower of zero for P R I s 2R 18 W. Then, because we must have a total power the entire circuit, we have P I P 12 W. V P R 30 18 T1.3
(a) The currents flowing downward through the resistances are v ab ab /R 1 and v ab ab /R 2. Then, the KCL equation for node a (or node b ) is v v I 2 I 1 ab ab R 1 R 2 Substituting the values given in the question and solving yields v aab b 8 V. (b) The power for current source I 1 is P I 1 v ab I 1 8 3 24 W .
Because the result is negative we know that energy is supplied by this current source. ab I 2 8 1 8 W . Because the The power for current source I 2 is P I 2 v result is positive, we know k now that energy is absorbed by this current source. 2 (c) The power absorbed by R 1 is P R 1 v ab / R 1 (8)2 / 12 5.33 W. The 2 power absorbed by R 2 is P R 2 v ab / R 2 (8) 2 / 6 10.67 W.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Full file at https://testbanku. https://testbanku.eu/Solution-Manualeu/Solution-Manual-for-Electrical-Engi for-Electrical-Engineering-7th-Edition neering-7th-Edition-by-Hambley-Chap -by-Hambley-Chapter-17-not-include ter-17-not-include
T1.4
2 0. Substituting values given in (a) Applying KVL, we have V s v 1 v the problem and solving we find v 1 8 V. (b) Then applying Ohm's law, we have i v 1 / R 1 8 / 4 2 A. (c) Again applying Ohm's law, we have R 2 v 2 / i 4 / 2 2 .
T1.5
Applying KVL, we have V s v x 0. Thus, v x V s 15 V. Next Ohm's law gives i x v x / R 15 / 10 1.5 A. Finally, KCL yields i sc i x av x 1.5 0.3 15 3 A.
T1.6
Applying Ohm’s law to the 40-Ω resistance, we have v 4 40 i 4 80 V. Since v 4 is also the voltage across the 20-Ω and 16-Ω resistances, we have i 3 v 4 / 16 80 / 16 5 A and i 2 v 4 / 20 80 / 20 4 A. Then applying KCL to the node joining the resistances, we have i 1 i 2 i 3 i 4 11 A. Then, applying Ohm’s law to the 10-Ω resistance, we have v i V. Finally, applying KVL, we have ha ve v v v V. 1
s
10 1 11 110 0
1
4
190 19 0
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